No. 01General Chemistry · Kinetics

Chapter 12 · Chemical Kinetics

Rate Laws
& Reaction Order

How concentration sets the speed of a reaction

CoverbookSHelf · §12.3–12.5
No. 02Contents

Outline — by the end of this section you will be able to

Objectives

  1. Express a reaction rate and read it off a concentration–time curve §12.1
  2. Write the rate law rate=k [A]m[B]n \text{rate}=k\,[\text{A}]^{m}[\text{B}]^{n} and name the orders §12.3
  3. Find each order from initial-rate data worked example
  4. Use integrated rate laws — and half-life — to track concentration over time §12.4
  5. Recognise the reaction order from the shape of the decay curve figure
Contents02 / 26
No. 03§12.1 · Reaction rates

§12.1 — What a rate measures

The rate of a reaction

For a balanced reaction aA+bB→cC+dD a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D} , the rate is a change in concentration per unit time — divided by the stoichiometric coefficient so every species reports the same number:

rate=−1aΔ[A]Δt=−1bΔ[B]Δt=+1cΔ[C]Δt. \text{rate} = -\frac{1}{a}\frac{\Delta[\text{A}]}{\Delta t} = -\frac{1}{b}\frac{\Delta[\text{B}]}{\Delta t} = +\frac{1}{c}\frac{\Delta[\text{C}]}{\Delta t}.

Always a positive number

Reactants disappear, so Δ[A]<0 \Delta[\text{A}]<0 ; the leading minus sign makes the reported rate positive.

Instantaneous rate = a slope

The rate at one instant is the slope of the tangent to the concentration–time curve at that moment.

Reaction rates03 / 26
No. 04§12.5 · Collision theory

§12.5 — Why reactions have a rate at all

The activation barrier

Molecules react only when they collide hard enough — and lined up right. The energy they must muster to reach the transition state is the activation energy Ea E_a .

Rate rises steeply with temperature

Only the fraction of collisions carrying at least Ea E_a can cross the barrier. Warming the sample enlarges that fraction — so the rate climbs fast (the Arrhenius picture).

Figure 1: The reacting system must pay Ea E_a to reach the transition state (‡), then falls to a lower-energy product — an exothermic step, ΔE<0 \Delta E<0 .

Collision theory04 / 26
No. 05§12.3 · Rate laws

§12.3 — The differential rate law

What a rate law says

The rate depends on concentration through a power law: rate=k [A]m [B]n. \text{rate} = k\,[\text{A}]^{m}\,[\text{B}]^{n}.

Orders are measured

The exponents m m and n n come from experiment — never from the balanced equation's coefficients.

k k is the rate constant

It sets the reaction's intrinsic speed and depends on temperature — but not on concentration.

Overall order is m+n m+n

The overall order sums the individual orders — and, as we'll see, fixes the units of k k .

The rate law05 / 26
No. 06§12.3 · Reaction order

§12.3 — What the order means

Order = how the rate answers concentration

The order in a reactant tells you exactly how the rate responds when you change its concentration:

Zero order — rate∝[A]0 \text{rate}\propto[\text{A}]^{0}

The rate ignores [A] [\text{A}] entirely; doubling the concentration changes nothing.

First order — rate∝[A]1 \text{rate}\propto[\text{A}]^{1}

Rate tracks concentration one-for-one; doubling [A] [\text{A}] doubles the rate.

Second order — rate∝[A]2 \text{rate}\propto[\text{A}]^{2}

Rate answers steeply; doubling [A] [\text{A}] multiplies the rate by 22=4 2^{2}=4 .

Reaction order06 / 26
No. 07§12.3 · Method of initial rates

§12.3 — Reading order from data

The method of initial rates

Method 12.3 — Change one concentration at a time

Run several trials, each changing a single reactant's starting concentration, and compare the initial rates. Between two trials that differ only in [A] [\text{A}] :

rate2rate1=([A]2[A]1)m. \frac{\text{rate}_2}{\text{rate}_1}=\left(\frac{[\text{A}]_2}{[\text{A}]_1}\right)^{m}.

Solve for the exponent m m — that is the order in A. Repeat, holding A fixed and varying B, to get n n .

Why it works: holding every other concentration constant isolates one reactant, so the rate ratio depends on that reactant's exponent alone.

Initial rates07 / 26
No. 08§12.3 · Order from a data set

§12.3 — Worked example

Worked example — find the rate law

Worked Example 12.3.1 — For A+B→products \text{A}+\text{B}\rightarrow\text{products}

Trial[A]0 [\text{A}]_0 (M)[B]0 [\text{B}]_0 (M)Initial rate (M/s)
10.100.102.0×10−3 2.0\times10^{-3}
20.200.108.0×10−3 8.0\times10^{-3}
30.200.201.6×10−2 1.6\times10^{-2}

Solution. Trials 1→2 double [A] [\text{A}] at fixed [B] [\text{B}] and the rate quadruples: 2m=4 2^{m}=4 , so m=2 m=2 . Trials 2→3 double [B] [\text{B}] at fixed [A] [\text{A}] and the rate doubles: 2n=2 2^{n}=2 , so n=1 n=1 . The rate law is rate=k [A]2[B] \text{rate}=k\,[\text{A}]^{2}[\text{B}] — third order overall. From Trial 1, k=2.0×10−3(0.10)2(0.10)=2.0 M−2s−1 k = \dfrac{2.0\times10^{-3}}{(0.10)^{2}(0.10)} = 2.0\ \text{M}^{-2}\text{s}^{-1} .

Order from data08 / 26
No. 09§12.3 · Your turn

§12.3 — Try it now

Find the order in each reactant

Try It Now 12.3.2 — For 2 NO+O2→2 NO2 2\,\text{NO}+\text{O}_2\rightarrow 2\,\text{NO}_2 , use the initial-rate data to find the order in NO, the order in O₂, and the overall order.

Trial[NO]0 [\text{NO}]_0 (M)[O2]0 [\text{O}_2]_0 (M)Initial rate (M/s)
10.0100.0102.5×10−5 2.5\times10^{-5}
20.0200.0101.0×10−4 1.0\times10^{-4}
30.0100.0205.0×10−5 5.0\times10^{-5}

Answer. Trials 1→2 double [NO] at fixed [O₂] and the rate rises ×4, so 2m=4 2^{m}=4 gives m=2 m=2 . Trials 1→3 double [O₂] at fixed [NO] and the rate rises ×2, so 2n=2 2^{n}=2 gives n=1 n=1 . The rate law is rate=k [NO]2[O2] \text{rate}=k\,[\text{NO}]^{2}[\text{O}_2] — third order overall.

Try it now · order09 / 26
No. 10§12.3 · The rate constant

§12.3 — Worked example

Compute k — and read off its units

Worked Example 12.3.3 — A reaction is first order in A and first order in B, so rate=k [A][B] \text{rate}=k\,[\text{A}][\text{B}] . In one trial [A]=0.050 M [\text{A}]=0.050\ \text{M} , [B]=0.10 M [\text{B}]=0.10\ \text{M} , and the initial rate is 1.5×10−3 M/s 1.5\times10^{-3}\ \text{M/s} . Find k and its units.

Solution. Solve the rate law for k and substitute:

k=rate[A][B]=1.5×10−3 M/s(0.050 M)(0.10 M)=0.30 M−1s−1. k=\frac{\text{rate}}{[\text{A}][\text{B}]}=\frac{1.5\times10^{-3}\ \text{M/s}}{(0.050\ \text{M})(0.10\ \text{M})}=0.30\ \text{M}^{-1}\text{s}^{-1}.

The units fall out of the algebra — M/sM⋅M=M−1s−1 \dfrac{\text{M/s}}{\text{M}\cdot\text{M}}=\text{M}^{-1}\text{s}^{-1} — exactly what an overall-second-order rate constant must carry.

Rate constant & units10 / 26
No. 11§12.3 · Your turn

§12.3 — Try it now

Your turn: find k and its units

Try It Now 12.3.4 — A reaction is first order in A: rate=k [A] \text{rate}=k\,[\text{A}] . When [A]=0.15 M [\text{A}]=0.15\ \text{M} the initial rate is 3.0×10−4 M/s 3.0\times10^{-4}\ \text{M/s} . What are k and its units?

Rearrange to k=rate/[A] k=\text{rate}/[\text{A}] before you reveal the answer.

Answer. k=3.0×10−4 M/s0.15 M=2.0×10−3 s−1 k=\dfrac{3.0\times10^{-4}\ \text{M/s}}{0.15\ \text{M}}=2.0\times10^{-3}\ \text{s}^{-1} . A first-order rate constant carries units of s−1 \text{s}^{-1} — no concentration in them, because the single [A] [\text{A}] cancels the M in the rate.

Try it now · k11 / 26
No. 12§12.4 · Integrated rate laws

§12.4 — From rate to concentration-in-time

Integrated rate laws

Integrating each differential law gives concentration as a function of time — and, for each order, a different quantity that plots as a straight line in t t :

OrderIntegrated lawLinear plotSlope
Zero[A]=[A]0−kt [\text{A}] = [\text{A}]_0 - kt [A] [\text{A}] vs t t −k -k
Firstln⁡[A]=ln⁡[A]0−kt \ln[\text{A}] = \ln[\text{A}]_0 - kt ln⁡[A] \ln[\text{A}] vs t t −k -k
Second1[A]=1[A]0+kt \dfrac{1}{[\text{A}]} = \dfrac{1}{[\text{A}]_0} + kt 1/[A] 1/[\text{A}] vs t t +k +k

Table 1: The three integrated rate laws in a single reactant, and the plot that linearises each.

The trick: whichever of [A] [\text{A}] , ln⁡[A] \ln[\text{A}] , or 1/[A] 1/[\text{A}] plots straight against time tells you the order directly.

Integrated rate laws12 / 26
No. 13§12.4 · First-order decay

§12.4 — Worked example

Track a first-order reaction over time

Worked Example 12.4.1 — A first-order reaction has k=0.025 s−1 k=0.025\ \text{s}^{-1} and starts at [A]0=0.80 M [\text{A}]_0=0.80\ \text{M} . What is [A] after t=60 s t=60\ \text{s} ?

Solution. Use the first-order integrated law ln⁡[A]=ln⁡[A]0−kt \ln[\text{A}]=\ln[\text{A}]_0-kt :

ln⁡[A]=ln⁡(0.80)−(0.025)(60)=−0.223−1.50=−1.72. \ln[\text{A}]=\ln(0.80)-(0.025)(60)=-0.223-1.50=-1.72.

Exponentiate both sides: [A]=e−1.72=0.18 M [\text{A}]=e^{-1.72}=0.18\ \text{M} . The reactant has fallen to about 22% 22\% of where it began.

First-order integrated law13 / 26
No. 14§12.4 · Your turn

§12.4 — Try it now

Your turn: how much is left?

Try It Now 12.4.2 — A first-order reaction has k=0.010 s−1 k=0.010\ \text{s}^{-1} and [A]0=1.00 M [\text{A}]_0=1.00\ \text{M} . Find [A] after t=120 s t=120\ \text{s} , then the fraction of A remaining.

Start from ln⁡[A]=ln⁡[A]0−kt \ln[\text{A}]=\ln[\text{A}]_0-kt , and note ln⁡(1.00)=0 \ln(1.00)=0 .

Answer. ln⁡[A]=0−(0.010)(120)=−1.20 \ln[\text{A}]=0-(0.010)(120)=-1.20 , so [A]=e−1.20=0.30 M [\text{A}]=e^{-1.20}=0.30\ \text{M} . The fraction remaining is e−kt=e−1.20=0.30 e^{-kt}=e^{-1.20}=0.30 , i.e. 30% 30\% .

Try it now · first order14 / 26
No. 15§12.4 · Order you can see

§12.4 — The shape of decay

The decay curve
tells the order

Release the same [A]0 [\text{A}]_0 three ways and the concentration falls in three unmistakable shapes.

Read the shape

Zero order falls in a straight line to zero; first order is an exponential with a constant half-life; second order is a hyperbola — steep at first, then a long tail.

Figure 2: Concentration–time decay for the three common orders, all starting from the same [A]0 [\text{A}]_0 .

The shape of decay15 / 26
No. 16§12.4 · Second-order decay

§12.4 — Worked example

A second-order reaction slows down

Worked Example 12.4.3 — A second-order reaction has k=0.50 M−1s−1 k=0.50\ \text{M}^{-1}\text{s}^{-1} and [A]0=0.10 M [\text{A}]_0=0.10\ \text{M} . Find [A] after t=40 s t=40\ \text{s} .

Solution. Use the second-order integrated law 1[A]=1[A]0+kt \dfrac{1}{[\text{A}]}=\dfrac{1}{[\text{A}]_0}+kt :

1[A]=10.10+(0.50)(40)=10+20=30 M−1. \frac{1}{[\text{A}]}=\frac{1}{0.10}+(0.50)(40)=10+20=30\ \text{M}^{-1}.

Invert: [A]=1/30=0.033 M [\text{A}]=1/30=0.033\ \text{M} . A second-order reaction fades ever more gently as it grows dilute — unlike the constant-percentage decay of first order.

Second-order integrated law16 / 26
No. 17§12.4 · Your turn

§12.4 — Try it now

Your turn: second-order decay

Try It Now 12.4.4 — A second-order reaction has k=0.20 M−1s−1 k=0.20\ \text{M}^{-1}\text{s}^{-1} and [A]0=0.50 M [\text{A}]_0=0.50\ \text{M} . Find [A] after t=25 s t=25\ \text{s} .

Reach for 1[A]=1[A]0+kt \dfrac{1}{[\text{A}]}=\dfrac{1}{[\text{A}]_0}+kt and invert at the end.

Answer. 1[A]=10.50+(0.20)(25)=2.0+5.0=7.0 M−1 \dfrac{1}{[\text{A}]}=\dfrac{1}{0.50}+(0.20)(25)=2.0+5.0=7.0\ \text{M}^{-1} , so [A]=1/7.0=0.14 M [\text{A}]=1/7.0=0.14\ \text{M} .

Try it now · second order17 / 26
No. 18§12.4 · Your turn

§12.4 — Try it now

Which plot gives a straight line?

Try It Now 12.4.5 — For a decomposition, a plot of 1/[A] 1/[\text{A}] against time is a straight line, while plots of [A] [\text{A}] and of ln⁡[A] \ln[\text{A}] against time both curve. What is the order, and what does the slope of the straight line equal?

Match the variable that plots straight to its integrated rate law.

Answer. A straight 1/[A] 1/[\text{A}] vs. t t line is the signature of a second-order reaction, 1[A]=1[A]0+kt \dfrac{1}{[\text{A}]}=\dfrac{1}{[\text{A}]_0}+kt ; the slope equals the rate constant k k (units M−1s−1 \text{M}^{-1}\text{s}^{-1} ). For reference: zero order makes [A] [\text{A}] linear, first order makes ln⁡[A] \ln[\text{A}] linear.

Try it now · linear plot18 / 26
No. 19§12.4 · Half-life

§12.4 — The half-life

Half-life

The half-life t1/2 t_{1/2} is the time for [A] [\text{A}] to fall to half its value. For a first-order reaction it is set only by k k :

t1/2=ln⁡2k. t_{1/2}=\frac{\ln 2}{k}.

The constant half-life test

Because t1/2 t_{1/2} contains no [A]0 [\text{A}]_0 , a first-order reaction takes the same time to halve, again and again — a fingerprint you can spot with no table of k k .

Figure 3: Equal-width half-lives each halve [A] [\text{A}] : [A]0→[A]0/2→[A]0/4→[A]0/8 [\text{A}]_0 \to [\text{A}]_0/2 \to [\text{A}]_0/4 \to [\text{A}]_0/8 .

Half-life19 / 26
No. 20§12.4 · Half-life arithmetic

§12.4 — Worked example

Worked example — a first-order half-life

Worked Example 12.4.6 — First-order decay with k=0.0693 s−1 k=0.0693\ \text{s}^{-1}

Find the half-life, then the time for [A] [\text{A}] to fall to one-eighth of its starting value.

Solution. t1/2=ln⁡2k=0.6930.0693=10 s t_{1/2}=\dfrac{\ln 2}{k}=\dfrac{0.693}{0.0693}=10\ \text{s} . Falling to 18 \tfrac{1}{8} is three successive halvings (12⋅12⋅12=18) \left(\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18\right) , and each takes one constant half-life — so the total time is 3 t1/2=30 s 3\,t_{1/2}=30\ \text{s} .

Half-life arithmetic20 / 26
No. 21§12.4 · Your turn

§12.4 — Try it now

Your turn: find the half-life

Try It Now 12.4.7 — A first-order reaction has k=1.5×10−3 s−1 k=1.5\times10^{-3}\ \text{s}^{-1} . Find its half-life t1/2 t_{1/2} , and confirm it does not depend on the starting concentration.

For first order, t1/2=ln⁡2k=0.693k t_{1/2}=\dfrac{\ln 2}{k}=\dfrac{0.693}{k} .

Answer. t1/2=0.6931.5×10−3=462 s t_{1/2}=\dfrac{0.693}{1.5\times10^{-3}}=462\ \text{s} (about 7.7 min). Because [A]0 [\text{A}]_0 cancels out of the first-order half-life, every successive half-life takes the same 462 s.

Try it now · half-life21 / 26
No. 22§12.4 · Counting half-lives

§12.4 — Worked example

Counting half-lives

Worked Example 12.4.8 — A first-order sample has a half-life of t1/2=30 min t_{1/2}=30\ \text{min} . What fraction of it remains after 2.0 hours?

Solution. Count how many half-lives fit the elapsed time: 2.0 h=120 min=4 t1/2 2.0\ \text{h}=120\ \text{min}=4\,t_{1/2} . Each half-life halves the amount, so the fraction remaining is

(12)4=116=0.0625=6.25%. \left(\tfrac{1}{2}\right)^{4}=\frac{1}{16}=0.0625=6.25\%.

After four half-lives only about 6% of the original reactant is left.

Counting half-lives22 / 26
No. 23§12.4 · Your turn

§12.4 — Try it now

Your turn: how much survives?

Try It Now 12.4.9 — An isotope decays by first-order kinetics with a half-life of t1/2=8.0 days t_{1/2}=8.0\ \text{days} . What fraction remains after 24 days?

First count how many half-lives fit into 24 days.

Answer. 24 days=3 t1/2 24\ \text{days}=3\,t_{1/2} , so the fraction remaining is (12)3=18=0.125=12.5% \left(\tfrac{1}{2}\right)^{3}=\dfrac{1}{8}=0.125=12.5\% .

Try it now · fraction left23 / 26
No. 24§12.4 · Order, units & half-life

§12.4 — What the order fixes

Units & half-life by order

Overall orderUnits of k k Half-life t1/2 t_{1/2}
ZeroM s−1 \text{M s}^{-1} [A]0/2k [\text{A}]_0 / 2k
Firsts−1 \text{s}^{-1} ln⁡2/k \ln 2 / k (constant)
SecondM−1s−1 \text{M}^{-1}\text{s}^{-1} 1/(k[A]0) 1 / (k[\text{A}]_0)

Table 2: How the overall order sets the rate-constant units and the half-life's dependence on concentration.

Read it: only the first-order half-life is independent of [A]0 [\text{A}]_0 — a constant t1/2 t_{1/2} is the signature of first-order decay.

Order & the constant24 / 26
No. 25Recap

Summary — the whole section on one page

What to carry away

Rate = k [A]m[B]n k\,[\text{A}]^{m}[\text{B}]^{n}

The orders m,n m,n are measured, not read from coefficients; m+n m+n is the overall order and fixes the units of k k .

Get order from data

Vary one concentration at a time and take the rate ratio ([A]2/[A]1)m \left([\text{A}]_2/[\text{A}]_1\right)^{m} ; or see which of [A] [\text{A}] , ln⁡[A] \ln[\text{A}] , 1/[A] 1/[\text{A}] plots straight in time.

Shape & half-life reveal order

A constant half-life t1/2=ln⁡2/k t_{1/2}=\ln 2/k — and an exponential decay curve — are the fingerprints of a first-order reaction.

Summary25 / 26
No. 26Colophon

Order, by
experiment.

A rate law is measured, not derived: find each order from how the rate answers a change in concentration, and the units, half-life, and decay curve all follow.

Next: §12.6 Reaction Mechanisms — how the rate law exposes the slowest elementary step.

bookSHelf · General Chemistry §12.3–12.5 Rate Laws, Order & Half-life

Adapted from OpenStax, Chemistry 2e (Chapter 12, Kinetics), published by Rice University under the Creative Commons Attribution 4.0 International (CC BY 4.0) license; the original is free at openstax.org. Figures 1–3 are dual-theme manim animations drawn for bookSHelf; deck styled in the notebook-tabs visual language (Open Design's fs-notebook-tabs template, MIT).

Colophon26 / 26